3.1070 \(\int \frac{c+d \tan (e+f x)}{(a+i a \tan (e+f x))^3} \, dx\)

Optimal. Leaf size=112 \[ \frac{d+i c}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac{x (c-i d)}{8 a^3}+\frac{-d+i c}{6 f (a+i a \tan (e+f x))^3}+\frac{d+i c}{8 a f (a+i a \tan (e+f x))^2} \]

[Out]

((c - I*d)*x)/(8*a^3) + (I*c - d)/(6*f*(a + I*a*Tan[e + f*x])^3) + (I*c + d)/(8*a*f*(a + I*a*Tan[e + f*x])^2)
+ (I*c + d)/(8*f*(a^3 + I*a^3*Tan[e + f*x]))

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Rubi [A]  time = 0.0832925, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {3526, 3479, 8} \[ \frac{d+i c}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac{x (c-i d)}{8 a^3}+\frac{-d+i c}{6 f (a+i a \tan (e+f x))^3}+\frac{d+i c}{8 a f (a+i a \tan (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*Tan[e + f*x])/(a + I*a*Tan[e + f*x])^3,x]

[Out]

((c - I*d)*x)/(8*a^3) + (I*c - d)/(6*f*(a + I*a*Tan[e + f*x])^3) + (I*c + d)/(8*a*f*(a + I*a*Tan[e + f*x])^2)
+ (I*c + d)/(8*f*(a^3 + I*a^3*Tan[e + f*x]))

Rule 3526

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^m)/(2*a*f*m), x] + Dist[(b*c + a*d)/(2*a*b), Int[(a + b*Tan[e + f*x])^(m + 1),
x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0]

Rule 3479

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(a + b*Tan[c + d*x])^n)/(2*b*d*n), x] +
Dist[1/(2*a), Int[(a + b*Tan[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && LtQ[n
, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{c+d \tan (e+f x)}{(a+i a \tan (e+f x))^3} \, dx &=\frac{i c-d}{6 f (a+i a \tan (e+f x))^3}+\frac{(c-i d) \int \frac{1}{(a+i a \tan (e+f x))^2} \, dx}{2 a}\\ &=\frac{i c-d}{6 f (a+i a \tan (e+f x))^3}+\frac{i c+d}{8 a f (a+i a \tan (e+f x))^2}+\frac{(c-i d) \int \frac{1}{a+i a \tan (e+f x)} \, dx}{4 a^2}\\ &=\frac{i c-d}{6 f (a+i a \tan (e+f x))^3}+\frac{i c+d}{8 a f (a+i a \tan (e+f x))^2}+\frac{i c+d}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac{(c-i d) \int 1 \, dx}{8 a^3}\\ &=\frac{(c-i d) x}{8 a^3}+\frac{i c-d}{6 f (a+i a \tan (e+f x))^3}+\frac{i c+d}{8 a f (a+i a \tan (e+f x))^2}+\frac{i c+d}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.828803, size = 150, normalized size = 1.34 \[ \frac{\sec ^3(e+f x) ((-27 c+3 i d) \cos (e+f x)+2 (6 i c f x-c+6 d f x-i d) \cos (3 (e+f x))-9 i c \sin (e+f x)+2 i c \sin (3 (e+f x))-12 c f x \sin (3 (e+f x))-9 d \sin (e+f x)-2 d \sin (3 (e+f x))+12 i d f x \sin (3 (e+f x)))}{96 a^3 f (\tan (e+f x)-i)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*Tan[e + f*x])/(a + I*a*Tan[e + f*x])^3,x]

[Out]

(Sec[e + f*x]^3*((-27*c + (3*I)*d)*Cos[e + f*x] + 2*(-c - I*d + (6*I)*c*f*x + 6*d*f*x)*Cos[3*(e + f*x)] - (9*I
)*c*Sin[e + f*x] - 9*d*Sin[e + f*x] + (2*I)*c*Sin[3*(e + f*x)] - 2*d*Sin[3*(e + f*x)] - 12*c*f*x*Sin[3*(e + f*
x)] + (12*I)*d*f*x*Sin[3*(e + f*x)]))/(96*a^3*f*(-I + Tan[e + f*x])^3)

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Maple [B]  time = 0.034, size = 203, normalized size = 1.8 \begin{align*} -{\frac{c}{6\,f{a}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) ^{3}}}-{\frac{{\frac{i}{6}}d}{f{a}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) ^{3}}}-{\frac{{\frac{i}{16}}\ln \left ( \tan \left ( fx+e \right ) -i \right ) c}{f{a}^{3}}}-{\frac{\ln \left ( \tan \left ( fx+e \right ) -i \right ) d}{16\,f{a}^{3}}}+{\frac{c}{8\,f{a}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) }}-{\frac{{\frac{i}{8}}d}{f{a}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) }}-{\frac{{\frac{i}{8}}c}{f{a}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}}-{\frac{d}{8\,f{a}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}}+{\frac{\ln \left ( \tan \left ( fx+e \right ) +i \right ) d}{16\,f{a}^{3}}}+{\frac{{\frac{i}{16}}\ln \left ( \tan \left ( fx+e \right ) +i \right ) c}{f{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c+d*tan(f*x+e))/(a+I*a*tan(f*x+e))^3,x)

[Out]

-1/6/f*c/a^3/(tan(f*x+e)-I)^3-1/6*I/f/a^3/(tan(f*x+e)-I)^3*d-1/16*I/f/a^3*ln(tan(f*x+e)-I)*c-1/16/f/a^3*ln(tan
(f*x+e)-I)*d+1/8/f/a^3/(tan(f*x+e)-I)*c-1/8*I/f/a^3/(tan(f*x+e)-I)*d-1/8*I/f/a^3/(tan(f*x+e)-I)^2*c-1/8/f/a^3/
(tan(f*x+e)-I)^2*d+1/16/f/a^3*ln(tan(f*x+e)+I)*d+1/16*I/f/a^3*ln(tan(f*x+e)+I)*c

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))/(a+I*a*tan(f*x+e))^3,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.5656, size = 217, normalized size = 1.94 \begin{align*} \frac{{\left (12 \,{\left (c - i \, d\right )} f x e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (18 i \, c + 6 \, d\right )} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (9 i \, c - 3 \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i \, c - 2 \, d\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{96 \, a^{3} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))/(a+I*a*tan(f*x+e))^3,x, algorithm="fricas")

[Out]

1/96*(12*(c - I*d)*f*x*e^(6*I*f*x + 6*I*e) + (18*I*c + 6*d)*e^(4*I*f*x + 4*I*e) + (9*I*c - 3*d)*e^(2*I*f*x + 2
*I*e) + 2*I*c - 2*d)*e^(-6*I*f*x - 6*I*e)/(a^3*f)

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Sympy [A]  time = 0.97146, size = 260, normalized size = 2.32 \begin{align*} \begin{cases} \frac{\left (\left (512 i a^{6} c f^{2} e^{6 i e} - 512 a^{6} d f^{2} e^{6 i e}\right ) e^{- 6 i f x} + \left (2304 i a^{6} c f^{2} e^{8 i e} - 768 a^{6} d f^{2} e^{8 i e}\right ) e^{- 4 i f x} + \left (4608 i a^{6} c f^{2} e^{10 i e} + 1536 a^{6} d f^{2} e^{10 i e}\right ) e^{- 2 i f x}\right ) e^{- 12 i e}}{24576 a^{9} f^{3}} & \text{for}\: 24576 a^{9} f^{3} e^{12 i e} \neq 0 \\x \left (- \frac{c - i d}{8 a^{3}} + \frac{\left (c e^{6 i e} + 3 c e^{4 i e} + 3 c e^{2 i e} + c - i d e^{6 i e} - i d e^{4 i e} + i d e^{2 i e} + i d\right ) e^{- 6 i e}}{8 a^{3}}\right ) & \text{otherwise} \end{cases} + \frac{x \left (c - i d\right )}{8 a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))/(a+I*a*tan(f*x+e))**3,x)

[Out]

Piecewise((((512*I*a**6*c*f**2*exp(6*I*e) - 512*a**6*d*f**2*exp(6*I*e))*exp(-6*I*f*x) + (2304*I*a**6*c*f**2*ex
p(8*I*e) - 768*a**6*d*f**2*exp(8*I*e))*exp(-4*I*f*x) + (4608*I*a**6*c*f**2*exp(10*I*e) + 1536*a**6*d*f**2*exp(
10*I*e))*exp(-2*I*f*x))*exp(-12*I*e)/(24576*a**9*f**3), Ne(24576*a**9*f**3*exp(12*I*e), 0)), (x*(-(c - I*d)/(8
*a**3) + (c*exp(6*I*e) + 3*c*exp(4*I*e) + 3*c*exp(2*I*e) + c - I*d*exp(6*I*e) - I*d*exp(4*I*e) + I*d*exp(2*I*e
) + I*d)*exp(-6*I*e)/(8*a**3)), True)) + x*(c - I*d)/(8*a**3)

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Giac [A]  time = 1.414, size = 189, normalized size = 1.69 \begin{align*} -\frac{\frac{6 \,{\left (i \, c + d\right )} \log \left (\tan \left (f x + e\right ) - i\right )}{a^{3}} + \frac{6 \,{\left (-i \, c - d\right )} \log \left (i \, \tan \left (f x + e\right ) - 1\right )}{a^{3}} + \frac{-11 i \, c \tan \left (f x + e\right )^{3} - 11 \, d \tan \left (f x + e\right )^{3} - 45 \, c \tan \left (f x + e\right )^{2} + 45 i \, d \tan \left (f x + e\right )^{2} + 69 i \, c \tan \left (f x + e\right ) + 69 \, d \tan \left (f x + e\right ) + 51 \, c - 19 i \, d}{a^{3}{\left (\tan \left (f x + e\right ) - i\right )}^{3}}}{96 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))/(a+I*a*tan(f*x+e))^3,x, algorithm="giac")

[Out]

-1/96*(6*(I*c + d)*log(tan(f*x + e) - I)/a^3 + 6*(-I*c - d)*log(I*tan(f*x + e) - 1)/a^3 + (-11*I*c*tan(f*x + e
)^3 - 11*d*tan(f*x + e)^3 - 45*c*tan(f*x + e)^2 + 45*I*d*tan(f*x + e)^2 + 69*I*c*tan(f*x + e) + 69*d*tan(f*x +
e) + 51*c - 19*I*d)/(a^3*(tan(f*x + e) - I)^3))/f